SOLUTION: Kumar walks around a rectangular field the length of which is twice its width. He then walks around another rectangular field half as wide but having the same perimeter as the firs

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Question 449030: Kumar walks around a rectangular field the length of which is twice its width. He then walks around another rectangular field half as wide but having the same perimeter as the first field. If the difference in area between the two fields is 432 m, find the length of the second field.
Answer by ankor@dixie-net.com(22740) (Show Source):
You can put this solution on YOUR website!
Kumar walks around a rectangular field the length of which is twice its width.
He then walks around another rectangular field half as wide but having the same perimeter as the first field.
If the difference in area between the two fields is 432 m, find the length of the second field.
:
Let x = width of original field
then
2x = length of original field
and
2x + 2(2x) =
6x = perimeter of the original field
and
2x^2 = area of original field
:
L = length of the 2nd field
and
.5x = width of the 2nd field
and
2L + 2(.5x) =
2L + x = perimeter of the 2nd field
and
.5x*L = area of the 2nd field
:
Perimeter the same, therefore:
2L + x = 6x
2L = 6x - x
2L = 5x
L = x
L = 2.5x
:
"If the difference in area between the two fields is 432 m,"
2x^2 - .5x*L = 432
Replace L with 2.5x
2x^2 - .5x(2.5x) = 432
2x^2 - 1.25x^2 = 432
.75x^2 = 432
x^2 =
x^2 = 576
x =
x = 24 m the width of the original width
2(24) = 48 m the length of the original field
then
L = 2.5(24)
L = 60 m is the length of the 2nd field
:
:
Check this
1st field 48 by 24, 2nd field: 60 by 12
Check solution by finding the perimeter of each
2(48) + 2(24) = 144
2(60) + 2(12) = 144, confirms our solution